Chapter
Matrix
Contents
1.1. Matrix Addition, Scalar Multiplication and Transposition
1.2. Matrix Multiplication
1.3. Matrix Inverses.
1.4. Determinants
1.1. Matrix Addition, scalar multiplication
and transposition
Definitions
⚫ An mxn matrix (or a matrix of size mxn) is a
rectangular array of numbers with m rows and
n columns
3 columns
 2 −3 5 
A=

1 0 4 
the (1,3)-entry of A
2 rows
A is a 2x3 matrix (or a matrix of size 2x3)
1.1. Matrix Addition, scalar multiplication
and transposition
The mxn matrix
⚫ The (i,j)-entry of A (denoted by aij) lies in row
i and column j
⚫ A is denoted simply as A=[aij] or A=(aij)
 a11
a
21

A=
 ...

 am1
a12
a22
...
am 2
a1n 

... a2 n 
... ... 

... amn 
...
2 refers to the row
n refers to the column
1.1. Matrix Addition, scalar multiplication
and transposition
Definitions
1/ An mxm matrix is called a square matrix (ma trận
vuông) of size m.
2/ The zero matrix (ma trận không) of size mxn (
denoted by (0mxn) is the matrix that its all entries are 0
3/ If A=[aij] is an mxn matrix then -A refers to the negative
matrix (ma trận đối) of A and defined by
-A=[-aij].
1.1. Matrix Addition, scalar multiplication
and transposition
4/ Identity matrices (ma trận đơn vị )
1 0 0 
1 0 
I 2 =   , I 3 = 0 1 0  ,...


0 1
0 0 1 
An identity matrix I is a square matrix with 1’s on the
main diagonal and zeros elsewhere
1.1. Matrix Addition, scalar multiplication
and transposition
5/ Triangular matrices
▪ Upper triangular matrix: all entries below and to
the left the main diagonal are zeros
▪ Lower triangular matrix: its tranposition is upper
triangle matrix, that means every entry above and to
the right the main diagonal is zero
▪ Matrix A is called triangular if it is upper or lower
triangular
▪ For example, every row-echelon matrix is upper
2 0 0 0
triangular
 2 0 0 0
 0 5 −3 0 



U =
0 0 0 0 L =  2 0 0 0


 −7 8 5 0 
0
0
0
0


1.1. Matrix Addition, scalar multiplication
and transposition
Two matrices are called equal if
⚫ They have the same size
⚫ Corresponding entries are equal
If A=[aij], B=[bij] then A=B means aij=bij for all i and j
1.1. Matrix Addition, scalar multiplication
and transposition
Example 1:
a b 
 1 0
1 2 −1
Given A = 
,B = 

 ,C = 
3 0 1 
b d 
 −1 2 
disscuss the possibility that A = B, B = C , A = C
1.1. Matrix Addition, scalar multiplication
and transposition
Matrix Addition of same size matrices
⚫ If A=[aij], B=[bij] then the sum matrix A+B is
defined by A+B=[aij+bij]
⚫ The difference A-B is a matrix defined by
A-B=A+(-B)=[aij-bij] for all mxn matrices A
and B
Note that A-A=0, A+0=A (0 is zero matrix) for
all mxn matrix A
1.1. Matrix Addition, scalar multiplication
and transposition
Example 2:
(
) (
)
−1 2 4
3 −2 6 , compute A + B
If A = 3 0 5 , B =
1 4 7
1
2
1.1. Matrix Addition, scalar multiplication
and transposition
Solution:
(
−1 2 4
A=
3 0 5
)
 2 0 10 
A+ B = 

 4 4 12 
(
3 −2 6
B=
1 4 7
)
1.1. Matrix Addition, scalar multiplication
and transposition
Example 3:
Find a, b and c if  a b c  + c a b  = 3 2 −1
1.1. Matrix Addition, scalar multiplication
and transposition
Solution:
 a b c + c a b = 3 2 −1
  a + c b + a c + b  = 3 2 −1
a + c = 3
a = 3


 b + a = 2  b = −1
c + b = −1 c = 0


1.1. Matrix Addition, scalar multiplication
and transposition
Properties
If A ,B and C are any matrices of the same size, then
⚫ A+B=B+A (commutative law: giao hoán)
⚫ A+(B+C)=(A+B)+C (associative law: kết hợp)
1.1. Matrix Addition, scalar multiplication
and transposition
Example:
3
1
2
0
Solve 
+ X =
 , where X is a matrix.
 −2 1 
 −1 2 
1.1. Matrix Addition, scalar multiplication
and transposition
Solution:
We have
3
1
2
0

+ X =

 −2 1 
 −1 2 
1
0  3
2   −2 −2 
 X =
−
=

 −1 2   −2 1  1 1 
 −2 −2 
So, X = 

1 1 
1.1. Matrix Addition, scalar multiplication
and transposition
Scalar Multiplication (phép nhân vô hướng)
⚫ Suppose A=[aij] is an mxn matrix and k is a
real munber, the scalar multiple kA is a
matrix defined by kA=[kaij]
⚫ kA=0→ (either k=0 or A=0)
⚫ (k=0 or A=0)→kA=0
Scalar: a quantity that has magnitude, but not direction; -- distinguished from a
vector, which has both magnitude and direction (Webster Dictionary) → One
value
1.1. Matrix Addition, scalar multiplication
and transposition
Example:
3 −1 4 
1 2 −1
If A = 
,B = 
 , compute
2 0 6
0 3 2 
a/ 5A
b/ 1 B
2
c/ 3 A − 2 B
1.1. Matrix Addition, scalar multiplication
and transposition
Theorem
Let A, B and C denoted arbitrary m  n matrices where m and
n are fixed. Let k and p denoted arbitrary real numbers. Then
1/ A + B = B + A
2/ A + ( B + C ) = ( A + B ) + C
3/ There is an m  n matrix 0, such that 0 + A = A for each A
4/ For each A there is an m  n matrix, − A such that
A + ( − A) = 0
5/ k ( A + B ) = kA + kB
7/ ( kp ) A = k ( pA)
6/ ( k + p ) A = kA + pA
8/ 1.A = A
1.1. Matrix Addition, scalar multiplication
and transposition
Transpose
⚫ If A=[aij] is any mxn matrix,the transpose of A, written
AT, is an nxm matrix defined by AT=[aji]
The row i of A is the column i of AT
⚫ The column j of A is the row j of AT
⚫
1.1. Matrix Addition, scalar multiplication
and transposition
Example:
Write down the transpose of each of the following
matrices
1 
 
a/ A =  3  , B = 5 2 6
 
 2 
1 2 


b/ C = 3 4 


 5 6 
1.1. Matrix Addition, scalar multiplication
and transposition
Solution:
1 
 
a/ A =  3   AT = 1 3 2
 
 2 
5 
 
B = 5 2 6  BT =  2 
 
 6 
Solution:
1.1. Matrix Addition, scalar multiplication
and transposition
1 2 


1 3 5 
T
b/ C = 3 4   C = 



2 4 6
 5 6 
1.1. Matrix Addition, scalar multiplication
and transposition
Theorem:
1/ (AT)T=A
2/ (kA)T=k(AT)
3/ (A+B)T=AT+BT
1.1. Matrix Addition, scalar multiplication
and transposition
Definitions
1/ If A=[aij] is any mxn matrix, then a11,a22,a33,…, are
called the main diagonal (đường chéo chính) of A
2/ If A=AT then A is called symmetric (đối xứng). In
this case, A is a square matrix
1.1. Matrix Addition, scalar multiplication
and transposition
Example 1:
 3 1 −1


D =  1 3 2  is symmetric matrix


 −1 2 1 
 3 1 −1


T
because D =  1 3 2  = D


 −1 2 1 
1.1. Matrix Addition, scalar multiplication
and transposition
Example 2:
a/ If A and B are symmetric n  n matrix, show that
A + B is symmetric.
b/ Suppose a square matrix A = 2 AT . Show that A = 0
1.2 Matrix Multiplication
Dot product (tích vô hướng)
Example:
 3
 
(1 2 3)   2  = 1  3 + 2  2 + 3  1 = 10
1
 
1.2 Matrix Multiplication
⚫ Suppose A=[aij] is an mxk matrix and B=[bij] is an kxn
matrix, then the product AB=[cij] is an mxn matrix
whose the (i,j)-entry is the dot product of row i of A
and column j of B
⚫ cij=(row i of A).(column j of B)
⚫ Note that AmxkBkxn is a mxn matrix
Example:
AB  BA
1 2

  1 0 −1 2 
 0 −1  


 1 2 1 0 
 −2 0 


1.2 Matrix Multiplication
Theorem 1:
Assume that k is an arbitrary scalar and A, B and C are
matrices of sizes such that the indicated can be performed
a/ IA = A, BI = B
b/ A ( BC ) = ( AB ) C
c/ A ( B + C ) = AB + AC , A ( B − C ) = AB − AC
d/ ( B + C ) A = BA + CA; ( B − C ) A = BA − CA
e/ k ( AB ) = ( kA) B = A ( kB )
T
f/ ( AB ) = BT AT
1.2 Matrix Multiplication
Example:
Suppose that A and B are nxn matrices. Simplify the
expression
A ( BC − CD ) + B ( C − D ) A + BDA
1.2 Matrix Multiplication
Solution:
We have A ( BC − CD ) + B ( C − D ) A + BDA
= ABC − ACD + BCA − BDA + BDA
= ABC − ACD + BCA
1.2 Matrix Multiplication
Example:
1 −1
a b
Let A = 
;B = 


0 0 
1 0
Find B such that B commutes with A and B 2 is zero
matrix.
3
5
1.2 Matrix Multiplication
Let square matrix Ann .
0
A = In
2
A = A. A
3
A = A. A. A

n
A = A. A.... A ( n times)
3
6
1.2 Matrix Multiplication
Example:
 1 3
Let A = 

0
1


2
3
, find A , A , A
200
 1 3  1 3   1 6   1 3  2 
A = A. A = 
=
=




0
1
0
1
0
1
0
1


 
 

2
 1 6  1 3   1 9   1 3  3 
A = A .A = 
=
=




0
1
0
1
0
1
0
1


 
 

3
2
So,
200
A
 1 3  200   1 600 
=
=


1  0 1 
0
3
7
1.2 Matrix Multiplication
Let f ( x ) = an x + an−1 x
and square matrix Ann
Then:
+ ... + a1 x + a0
f ( A ) = an A + an−1 A
+ ... + a1 A + a0 I n
n
n
n −1
n −1
where: I n is identity matrix
3
8
1.2 Matrix Multiplication
Example:
 2 −1
2
Cho A = 
; f ( x ) = 2 x − 4 x + .3

3 4 
Compute f ( A) .
3
9
1.2 Matrix Multiplication
Example:
Let
 2 −1
2
A=
; f ( x) = 2x − 4x + 3

3 4 
Compute f ( A) .
2
Solution: We have, f ( A ) = 2 A − 4 A + 3I 2
 2 −1  2 −1  2 −1  1 0 
f ( A) = 2 

− 4
+ 3




3 4  3 4  3 4  0 1
 1 −6   8 −4   3 0 
f ( A) = 2 
−
+



18 13  12 16   0 3 
 −3 −8 
Hence, f ( A ) = 

 24 13 
1.2 Matrix Multiplication
Row-echelon matrix
A matrix is said to be in row-echelon form ( and will be called a rowechelon matrix) if it satisfiles the following three conditions:
1/ All zero rows ( consisting entirely zeros) are at the bottom.
2/ The first nonzero entry from the left in each nonzero row is a 1,
called the leading 1 for that row.
3. Each leading 1 is to the right of all leading 1s in the rows above it.
A row-echelon matrix is said to be in reduced row-echelon form
(and will be called a reduced row-echelon matrix) if, in addition, it
satisfies the following condition:
4/ Each leading 1 is the only nonzero entry in its column.
1.2 Matrix Multiplication
Row-echelon matrix
Example 1:
a/
b/
is not row-echelon matrix
is not row-echelon matrix
1.2 Matrix Multiplication
Row-echelon matrix
Example 2:
a/
b/
is row-echelon matrix
is row-echelon matrix
1.2 Matrix Multiplication
Row-echelon matrix
Definition:
The following operations, called elementary operations
1/ Interchange two rows.
ri  rj
2/ Multiply one row by a nonzero number.
ri →  .ri (  0 )
3/ Add a multiple of one row to a different row.
ri → ri +  .rj
1.2 Matrix Multiplication
Row-echelon matrix
Example 1: Use the elementary operations, find the rowechelon matrix
0 0 2
A = 1 2 1 


 0 −1 1 


Step 1: Find leading 1
0

A= 1
0

2

2
1 
−1 1 
0
4
5
1.2 Matrix Multiplication
Row-echelon matrix
Step 2:
0

A=1

0

Step 3:
1

0

0

0
2
−1
2
0
−1
1
2
 r1 r2 
1  →0


0
1 

1
1
 r3 r2 
2 →0

0

1 

2
0
−1
2
−1
0
1

2

1 
1

1
2 

1.2 Matrix Multiplication
Row-echelon matrix
Example 2:
Use the elementary operations, find the row- echelon matrix
 1
 2
A=
 3

 −1
1 −1
2 1

3 −1 4 5

2 −3 7 4 

1 2 −3 1 
1.2 Matrix Multiplication
Row-echelon matrix
Solution:
Step 1:
 11
 2
A=
 3

 −1
r2 →r2 − 2 r1
1 −1
2 1

3 −1 4 5

2 −3 7 4 

1 2 −3 1 
 1 1 −1 2 1 
r3 →r3 −3 r1
 0 1

1
0
3
r4 →r4 + r1

⎯⎯⎯⎯
→
 0 −1 0 1 1 


1 −1 2 
0 2
1.2 Matrix Multiplication
Khi khử phần tử
leading trên cột
2 không được sử
dụng hàng 1
Row-echelon matrix
Step 2:
 1 1 −1 2 1 
 0 1

1
0
3
1


 0 −1 0 1 1 


1 −1 2 
0 2
 1
r3 →r3 + r2
 0
r4 →r4 − 2 r2
⎯⎯⎯⎯
→
 0

0
1
1
0
0
−1
1 
1 0 3 

1 1 4 

− 1 −1 − 4 
2
1.2 Matrix Multiplication
Step 3:
1 1
 0 11

 0 0

0 0
−1
1 

1 0 3

11 1 4 

− 1 −1 − 4 
 1 1 −1
 0 1 1
r4 →r4 + r3
⎯⎯⎯⎯
→
 0 0 1

0 0 0
2
Khi khử phần tử
leading trên cột 3
không được sử dụng
hàng 1 và hàng 2.
Tương tự cho các
cột tiếp theo ……
2 1
0 3

1 4

0 0
1.2 Matrix Multiplication
Example 3:
Use the elementary operations, find the row- echelon matrix
1 1 1 3
1 1 2 3




a/ A = 2 5 7 2 b/ B = 2 5 7 2




1 2 3 1 
1 8 3 1




 3 −5 4 


c/ C = 2 −1 3


1 2 5


 6 0 4


d/ D = 2 6 8


 −3 4 1 


1.3 Matrix Inverse
Definition
▪ If A is a square matrix, a matrix B is called an
inverse (nghịch đảo) of A if and only if AB=I and
BA=I.
▪ A matrix A that has an inverse is called an invertible
(khả nghịch) matrix.
▪
Denoted B = A−1
1
Noted: A 
A
−1
1.3 Matrix Inverse
Example 1:
2
Let A = 
5

We have
1
 3 −1
and B = 


3
−
5
2


 2 1  3 −1  1 0 
AB = 
=



 5 3  −5 2   0 1 
 3 −1 2 1   1 0 
BA = 
=



 −5 2  5 3   0 1 
So,
 3 −1
A = B=

−
5
2


−1
1.3 Matrix Inverse
Theorem:
If B and C are both inverses of A, then B = C
1.3 Matrix Inverse
Example 2:
If the matrix A satisfy A2 − 4 A + 3I = 0 , then A−1 is
a/ exists but not enough information to find.
1
b/ ( 4 I − A )
3
c/
1
A − 4I )
(
3
d/ not exits the inverse of A.
1
e/ ( 4 − A )
3
1.3 Matrix Inverse
Solution:
We have
A − 4 A + 3I = 0
2
A2 − 4 A + 3 I = 0
 4 A − A = 3I
 4 A − A2 = 3 I
1
 ( 4 A − A2 ) = I
3
1

  ( 4 I − A )  A = I ( *)
3

From (*) and (**), we have
1
 ( 4 A − A2 ) = I
3
1

 A  ( 4 I − A )  = I (**)
3

2
1
A = ( 4I − A)
3
−1
1.3 Matrix Inverse
Example 3:
Show that A3=I and find A-1 if
0 −1
A=

1
−
1


1.3 Matrix Inverse
Solution:
We have
 −1 1 
A = AA = 

−
1
0


2
 −1 1  0 −1 1 0
A = A A=
=
= I2




 −1 0 1 −1 0 1 
3
2
And, A3= A2A=AA2=I2
▪ So, A-1=A2
▪
1.3 Matrix Inverse
Matrix Inversion Algorithm
elementary row operations
 A I − − − − − − − → I A 
input
Theorem 3.
Either any square matrix can be reduced to I or not.
In the first case, the algorithm produces A-1;
in the second, A-1 does not exist.
-1
output
1.3 Matrix Inverse
Example
Find the inverse of the matrix
 1 0 −1


A= 3 2 0


 −1 −1 0 
1.3 Matrix Inverse
Example:
 1 0 −1 1 0 0 
 3 2 0  0 1 0 



 −1 −1 0  0 0 1 
1

r r

→
0

0
1 0
−r
→ 0 1

0 2
2
2
3
1 0 −1  1 0 0 
0 2 3   −3 1 0 



0 −1 −1  1 0 1 
−3 r1 + r2
r1 + r3
→
0 −1  1 0 0 



−1 −1 1 0 1


2 3   −3 1 0 
−1  1 0 0 
−2 r + r
1   −1 0 −1 →


3   −3 1 0 
2
3
1 0 −1  1 0 0 
0 1 1   −1 0 −1



0 0 1   −1 1 2 
1.3 Matrix Inverse
Example
− r3 + r2
r3 + r1
→
1 0 0   0 1 2 
0 1 0   0 −1 −3



0 0 1   −1 1 2 
0 1 2
A−1 =  0 −1 −3


 −1 1 2 
1.3 Matrix Inverse
Properties
Theorem 4
1/ (A-1)-1=A
2/ (AB)-1=B-1A-1
(A1A2…Ak)-1=Ak-1…A2-1A1-1
3/ (AT)-1=(A-1)T
4/ (Ak)-1=(A-1)k
5/ (kA)-1=A-1/k
6/ I-1=I
Corollary
A square matrix A is invertible iff AT is invertible
1.3 Matrix Inverse
Example
Find A if
( A − 2I )
T
−1
 2 1
=

−1 0
1.3 Matrix Inverse
Solution:
We have
(
A − 2I = ( A − 2I )
T
T
)
−1 −1
−1
 2 1
0 −1
=
=


−
1
0
1
2




1.3 Matrix Inverse
Solution:
We have
(
A − 2I = ( A − 2I )
T
T
Hence
0 −1
A = 2I + 

1
2


T
So
2
A=
 −1
1
4 
)
−1 −1
−1
 2 1
0 −1
=
=


−
1
0
1
2




1.3 Matrix Inverse
Theorem
The following conditions are equivalent for an nxn
matrix A
▪ 1. A is invertible
▪ 2. The homogeneous system AX=0 has only the
trivial solution X=0
▪ 3. A can be carried to In by elementary row
operations
▪ 4. The system AX=B has unique solution for every
choice of column B
▪ 5. There exist an nxn matrix C such that AC=In
1.3 Matrix Inverse
Theorem
Suppose AX=B is a system of n equations in n
variables and A is an invertible matrix. Then the
system has the unique solution
X=A-1B
Similar:
XA=B  X=BA-1
1.3 Matrix Inverse
Example 1:
0 −1
Let A = 

1
−
1


Solve the system AX=B with B=[1 2]T
1.3 Matrix Inverse
Solution:
AX = B  A−1 AX = A−1B
 −1 1  1   1 
 X =A B=
= 



 −1 0   2   −1
−1
1.3 Matrix Inverse
Example 2: Let
 1 −2 3 
3 2 2
 2 −1 3 
;B =  4 5 0  ; C =  1 1 1 


A= 0 3 1






0 1 2
1 0 1 
 5 −2 4 






Find X such that:
a/ C. X = B
b/ X .C = B
c/ A. X = B
d/ X . A = B
7
1
1.3 Matrix Inverse
Solution:
 −7 −12 3 
a/ CX = B  X = C −1B =  4
4 −2 

 7 13 −1 


 −2 2 5 
b/ X.C = B  X = B.C −1 =  4 −3 −5 


 −2 5 1 


7
2
1.3 Matrix Inverse
Solution:
c/
−1 
 −3 / 11 24 / 11
A.X = B  X = A −1 B =  23 / 22 47 / 22 −1/ 2 


 19 / 22 −31/ 22 3 / 2 


d/
 41/ 22 −9 / 22 −6 / 11
X.A = B  X = B A −1 =  −81/ 22 43 / 22 25 / 11 


 25 / 22 9 / 22 −5 / 11 


1.3 Matrix Inverse
Corollary
If A and C are square matrices such that AC=I,
then also CA=I. In particular, both A and C are
invertible, C=A-1 and A=C-1.
1.4. Determinants
The Cofactor Expansion
If A=[a] then the determinant of
A, denoted by detA = a
a
▪ If A is an 2x2 matrix then A = 
▪
det A =
▪
a b
c d
b

c d 
= ad − bc
If A is an 3x3 matrix then the
determinant of A is defined by
a b
A =  −7  ;det A = −7
c
det A = d
e
f = a. ( + ) .
g
h
i
e
f
h
i
+ b. ( − ) .
d
f
g
i
+
a
A = d

 g
-
+ c. ( + ) .
d
e
g
h
+
c
f

i 
b
e
h
1.4. Determinants
The determinant of 3x3 matrix
a b
c
det A = d
e
f = a. ( + ) .
g
h
i
e
f
h
i
+ b. ( − ) .
d
f
g
i
+ c. ( + ) .
d
e
g
h
= aei + bfg + cdh −.ceg − afh − bdi
1.4. Determinants
The determinant of 3x3 matrix
col1 col 2 col 3 col1 co
coll12 col 2 col 3 col1 col 2
1b
−5
4
1
−5
a
b
c
a
d
e
f
d
0e
2
6
0
2
g
h
i
g
0h
0
3
0
0
-
-
+
+
-
det A = aei + bfg + cdh − ceg − afh − bdi
+
1.4. Determinants
Find detA if
1 −5 4 
A = 0 2 6 


0 0 3 
col1 col 2 col 3 col1 col 2
1
−5
4
1
−5
0
2
6
0
2
0
0
3
0
0
- + + +
Note that : only use with 3x3 matrices
1.4. Determinants
1
0
A=
0

0
2 −1 5 
6 4 3

7 −1 0 

1 8 2
( −1)1+1 .det ( A11 )

6
4
a

11
3
0
4
3
det A = 1 . ( +1) . 7 −1 0 + 2. ( −1) . 0 −1 0 +
1
0 6 3
( −1)( +1) 0
8
2
0
0 6
4
7 0 − 5 ( −1) 0 7 −1
1 2 2
0 1
8
8
2
1.4. Determinants
Definition
Example:
1 2 −1 5
6 4 0
0 6 4 0
= 1 7 −1 0 = −68
0 7 −1 0
1 8 2
0 1 8 2
det A = a11.c11 + a21.c21 + a31.c31 + a41.c41
= a11.c11 = c11 = ( −1) .det A11 = det A11 =
2
= a13 .c13 + a23 .c23 + a33 .c33 = 2. ( −1)
3+ 3
.
6
4
7 −1
= 2. ( −6 − 28 ) = −68
1.4. Determinants
The (i,j)-cofactor (phần phụ đại số)
⚫ If A is an mxm matrix then the (i,j)-cofactor of A is
defined by
cij(A)=(-1)i+jdet(Aij)
⚫ Aij is the (m-1)x(m-1) matrix obtained from A by
deleting row i and column j of A
1.4. Determinants
The (i,j)-cofactor (phần phụ đại số)
c23(A)=(-1)2+3det(A23)=-14
Example:
1
0
A=
0

0
2 −1 5 
6 4 3

7 −1 0 

1 8 2
1 2 5 


A23 = 0 7 0


0 1 2 
1 −1 5
c32 = ( −1)
3+ 2
.det ( A32 ) = − 0
4
3 = − ( 8 − 24 ) = 16
0
8
2
1.4. Determinants
Definition
a b
det A = d
e
g
h
c
e
f
d
f = a . ( + ) .
+ b . ( − ) .
h i a
g
a

i
11
12
(1,1) − cofactor
f
d
e
+ c . ( + ) .
i a
g h

(1,2) − cofactor
13
(1,3) − cofactor
If A is an nxn matrix then the determinant of A is
defined by
⚫ detA=|A|=ai1ci1(A)+ai2ci2(A)+…+aimcim(A)
⚫ or detA=|A|= a1jc1j(A)+a2jc2j(A)+…+amjcmj(A)
1.4. Determinants
Properties
1/ If A has one row (or columm) of zeros then detA=0
2/ If A has two identical rows(columns) then detA=0
a b c
x y
z =0
a b c
3/ If A is an triangular matrix then detA is the product
of the entries on the main diagonal
3 2 −1 5
6 4 3
0 6 4 3
−1 0
= 3. 0 −1 0 = 3.6.
= 3.6. ( −1) .2
0 0 −1 0
0 2
0 0 2
0 0 0 2
Upper triangular
1.4. Determinants
Example:
Suppose A is an triangular matrix, a11=4 and c11(A)=11.
What is the product of all entries on the maindiagonal ?
A)11
B)44
C)-11
D)-44
a11
0

A= 0

 ...
 0
a12 ... ... a1n 
a22 ... ... a2n 
0 ... ... 0 

... ... ... ... 
0 ... ... ann 
Upper triangular
detA=a11c11(A) + a21c21(A) + … = a11c11(A)
1.4. Determinants
Determinants and elementary operations
a/ If B obtained from A by interchanging two rows (or
columns) then detB = - detA
Example: detC’= - detC=12
1
0
C=
0

0
2 −1 5 
6 4 3

0 −1 0 

0 0 2
1
0
C'= 
0

0
0
3 −2
1
2
5
−3 1
1
r1  r2
=
1
− 0
2
2 −1 5 
6 4 3

0 0 2

0 −1 0 
5
3 −2
−3 1
1
1.4. Determinants
Determinants and elementary operations
2/ If two rows (or columns) of a matrix is the same then
the determinant is zero.
Examples:
1
0
C=
1

1
2 −1 5 
6 4 3

2 5 6

2 5 6
1
0
C'= 
1

1
2 −1 5 
6 4 3
=C
2 5 6

2 5 6
det C = − det C ' = − det C  det C = 0
1.4. Determinants
Determinants and elementary operations
3/ If B is the matrix obtained from A by multiplying
one row (or cloumn) by a nonzero number k then
detB=kdetA.
Example:
detA’=35, detA=2detA’=2.35=70
1 0
0
0 7
0 = 2 0 7 0 = 70
2 4 10
1 0 0
1 2 5
1.4. Determinants
Examples: Find a such that
2
−2
8 −4
4 = a 2 −1
1 ?
0
4
1
4
a) 4
−8
8
8
1
0
b) 12
2
c) 48


  k.
Note that k . 





k matrix
d)64
k number
Note that a|A| means number a multiplies number |A|
while aA means number a multiplies matrix A.
1.4. Determinants
Determinants and elementary operations
4/ If A is a square matrix that one row(column) is a
multiple of another row(column) then the determinant
of A is zero
a b c
a b c
x
y
r3 = kr1
z = k x
ka kb kc
y
r1 = r3
z =0
a b c
1.4. Determinants
Determinants and elementary operations
5/ If B is the matrix obtained from A by adding a
multiple of one row (or column) to another row (or
column) then detB=detA
Example:
1 2
−3
0 7
0
2 4 −10
−2 r1 + r3
1 2 −3
= 0 7
0 = 1.7. ( −4 ) = −28
0 0 −4
1.4. Determinants
Examples
0
2 −1 9
2
2 −4 6
2
2 −4 6 r  r 0
= −
2 −2 1
3
2 −1 9
3
1
−3 4
−3 4
0
1
1
−2
3
0 −1
4
−8
0
2
−1
9
0
7
−4
9
1
−2
3
0 −1
4
−8
0
0
1
−1
0
0
0
−23
= − ( −2 )
1
= 2.7
2 −2 1
2
r2  r3
−24 r3 + r4
2
2
2 r2 + r3
7 r2 + r4
= 2
= −2
1
1 −2 3
0
2 −1 9
3
2 −2 1
−3 4
0
1
−2
3
0 −1
4
−8
0
0
7
−7
0
0
24 −47
1
= 2.7.1. ( −1) .1. ( −23 )
1 2 −1 5
0 6 4 3
Do your self: Find
1 −3 4 6
1
2
4
5
2
−3 r1 + r3
3 r1 + r4
= −2
0
1
−2
3
0
2
−1
9
0 −1
4
−8
0
−4
9
7
1
−2
3
0 −1
4
−8
0
0
1
−1
0
0
24 −47
1
= 2.7
1
1.4. Determinants
Theorem
1/ detA=det(AT)
2/ det(AB)=detA.detB
Note that
det(Ak)=(detA)k
det(A+B) ≠ detA+detB
3/ If A is an nxn matrix then det(kA)=kndetA.
4/ detA-1=1/detA.
1.4. Determinants
Example 1:
Suppose A is a nxn matrix, and k is a scalar. Which of the
following is(are) false ?
a) det(AT)=detA
b) det(AB)=detAdetB
c) det(A+B)=detA+detB
d) det(kA)=kdetA
e) det(kA)=kndetA
1.4. Determinants
Example 2:
Let A and B are nxn matrices. Which statement is false ?
a) det(A-1BA)=detB
b) det(AB-1A-1B)=1
c) detAdetA-1=1
d) detAdetAT=1
e) det(ATB)=det(BTA)
1.4. Determinants
Example 3:
a b c 


Let A =  p q r  and assume that det ( A) = 3 .


 u v w
 4u 2a − p 
Compute:


−1
a/det ( 2B ) where B =  4v 2b −q 


 4 w 2c −r 
1.4. Determinants
Example 3:
a b c 


Let A =  p q r  and assume that det ( A) = 3 .


 u v w
2 p −a + u 3u 

Compute:


b/ det ( 2C −1 ) where C =  2q −b + v 3v 


 2r −c + w 3w
1.4. Determinants
Determinant and Matrix Inverses
1/ If A is invertible then A→Identity matrix
2/ If detA ≠0 and A→B by elementary operations then
detB ≠0
3/ A is invertible if detA≠0
4/ Give a formula to find A-1.
1.4. Determinants
Determinant and Matrix Inverses
Example 1:
Use determinants to find which real values of c make
each of the following matrices invertible.
 0 c −c 
 1 0 3




a/ A =  3 −4 c 
b/ B =  −1 2 1 




 c −c c 
 2 5 8 
 c 1 0


c/ C =  0 2 c 


 −1 c 5 
1.4. Determinants
Determinant and Matrix Inverses
Example 2:
Use determinants to find which real values of c make
each of the following matrices invertible.
1 2 −1
4 c 3




a/ A =  c 2 c 
b/ B =  0 −1 c 




 2 c 1 
 5 c 4 
1 c −1


c/ C =  c 1 1 


0 1 c 
1.4. Determinants
(i,j)-Cofactor (phần bù đại số)
⚫ cij=cij(A)=(-1)i+jdet(Aij)
( it is a number)
⚫ Aij is the matrix obtained from A by deleting
row i and column j
1 2 −1 5 
0 Che
Che
6 dòng
4 2 3

cột
A=
0 7 −31 0 


0
1
8
2


1 2 5 


A23 = 0 7 0


0 1 2 
c23 = c23 ( A ) = ( −1)
1 2 5
2+3
det A23 = − 0 7 0 = −14
0 1 2
1.4. Determinants
Adjugate matrix
The adjugate matrix of A is the matrix
 c11
c
adjA =  12
 ...

Example 1:
c1n
c21 ... cn1 
c22 ... cn 2 
... ... ... 

c2 n ... cnn 
1 −2 0 
−1
−1
1 +1 3
1+2 1
A = 1 3 −1 . We have c11 = ( −1)
= 3, c12 = ( −1)
= −3,
0 1
2 1
2 0 1 
c11 = 3, c12 = −3, c13 = −6
 3 2 2
c21 = 2, c22 = 1, c23 = −4,
 adjA =  −3 1 1 
c31 = 2, c32 = 1, c33 = 5
 −6 −4 5
1.4. Determinants
Example:
Find the adjugate of each of the following matrices.
1 −1 2 
 5 1 3




a/ A =  −1 2 3
b/B =  3 1 0 




0 −1 1 
 1 4 8
 1 0 −1


c/C =  −1 1 0 


 0 −1 1 
 −1 2 2 

1
d/ D =  2 −1 2 
3

 2 2 −1
1.4. Determinants
Theorem of Adjugate Formula:
If A is any square matrix, then
⚫ A(adjA)=(detA)I
⚫ In particular, if detA≠0 then A is invertible and
1
A =
adjA
det A
−1
Example:
1 −1 2 
2 1 −3
A = 0 2 −1  det A = 2 and adjA= 0 1 1 
0 0 1 
0 0 2 
2 1 −3 1 1 / 2 −3 / 2 
1
 A−1 = 0 1 1  = 0 1 / 2 1 / 2 
2
0 0 2  0 0
1 
1.4. Determinants
Theorem of Adjugate Formula
Note:
1
A =
.adjA
det A
 adjA = ( det A ) . A−1
−1
 adjA = ( det A ) . A
−1
= ( det A ) .det A = ( det A )
n
−1
n −1
1.4. Determinants
Example:
Let A be a 2x2 matrix with detA=3.
Find det(adjA) ?
a) 1/3
b)1/2
c)2
d) 3
1.4. Determinants
Example:
Determine whether the statements is true or false.
1/ The determinant of a square matrix equal to the sum of all
entries of this matrix.
2/ The determinant of a square matrix is a matrix with same
size.
3/ The determinant of a square invertible matrix always
eaqual to 1.
4/ The determinant of a square matrix is 0 if it is not
invertible.
5/ By elementary operations, we can find the determinant of a
square matrix.
1.4. Determinants
Example:
6/ Suppose A, B, C are square matrices satisfying A=BC and
A is not invertible. Then B or C is not invertible.
7/ Suppose A, B, C are square matrices satisfying A=BC and
A is invertible. Then B and C are invertible.
8/ If AB=AC, then B=C.
9/ If AB=0 then A=0 or B=0.
10/ If A is invertible, then Ak is invertible for all positive
integer k.